$\overline{AB}$ = $17$ $\overline{BC} = {?}$ $A$ $C$ $B$ $17$ $?$ $ \sin( \angle BAC ) = \dfrac{15}{17}, \cos( \angle BAC ) = \dfrac{8}{17}, \tan( \angle BAC ) = \dfrac{15}{8}$
$\overline{AB}$ is the hypotenuse $\overline{BC}$ is opposite to $\angle BAC$ SOH CAH TOA We know the hypotenuse and need to solve for the opposite side so we can use the sine function (SOH) $ \sin( \angle BAC ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\overline{BC}}{\overline{AB}}= \frac{\overline{BC}}{17} $ Since we have already been given $\sin( \angle BAC )$ , we can set up a proportion to find $\overline{BC}$ $ \sin( \angle BAC ) = \dfrac{15}{17} = \frac{\overline{BC}}{17}$ Simplify. $\overline{BC} = 15$